Does a closed solution of this integral exist I saw a problem on Facebook with this integral except it was a definite integral from -2 to +2 and the answer was 0 since the function was odd. I am wondering if a closed solution exists or if this can even be integrated, its not an obvious integral as trig. Substitution would not work due to the $x$ term being present in the cosine function and outside. Is there perhaps a way to transform this to something a bit manageable?
$$ \int x^3 \cos \bigg( \frac{x}{2} \bigg) \sqrt{4-x^2}\, dx$$
 A: Let $x=2\sin u$ ,
Then $dx=2\cos u~du$
$\therefore\int x^3\cos\dfrac{x}{2}\sqrt{4-x^2}~dx$
$=16\int\sin^3u\cos^2u\cos\sin u~du$
$=16\int\sin^3u(1-\sin^2u)\cos\sin u~du$
$=16\int\sin^3u\cos\sin u~du-16\int\sin^5u\cos\sin u~du$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+3}u}{(2n)!}~du-16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+5}u}{(2n)!}~du$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+4}u}{(2n)!}~d(\cos u)-16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+2}u}{(2n)!}~d(\cos u)$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\cos^2u)^{n+2}}{(2n)!}~d(\cos u)-16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\cos^2u)^{n+1}}{(2n)!}~d(\cos u)$
$=16\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^kC_k^{n+2}\cos^{2k}u}{(2n)!}~d(\cos u)+16\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^kC_k^{n+1}\cos^{2k}u}{(2n)!}~d(\cos u)$
$=16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^k(n+2)!\cos^{2k+1}u}{(2n)!k!(n-k+2)!(2k+1)}+16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^k(n+1)!\cos^{2k+1}u}{(2n)!k!(n-k+1)!(2k+1)}+C$
$=16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^k(n+2)!\left(1-\dfrac{x^2}{4}\right)^k\sqrt{1-\dfrac{x^2}{4}}}{(2n)!k!(n-k+2)!(2k+1)}+16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^k(n+1)!\left(1-\dfrac{x^2}{4}\right)^k\sqrt{1-\dfrac{x^2}{4}}}{(2n)!k!(n-k+1)!(2k+1)}+C$
$=8\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^k(n+2)!(4-x^2)^k\sqrt{4-x^2}}{4^k(2n)!k!(n-k+2)!(2k+1)}+8\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^k(n+1)!(4-x^2)^k\sqrt{4-x^2}}{4^k(2n)!k!(n-k+1)!(2k+1)}+C$
